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prove that
one over square root of two pi times the integral of e raise to negative one half x squared, dx from positive infinity to negative infinity is equal to one.
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e raise to the negative power of half x squared will lead to 1 as x goes to infinity because negative half of x squared will become zero, e raised to the power of zero is one....so the answer should be one over square root of two pi.
i think thats correct if i had not mistaken the question
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http://z8.invisionfree.com/DYK/index.php?showtopic=136
So
If you use the substitution
you should get the answer to your question.
Last edited by JaneFairfax (2007-12-01 10:51:02)
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